Continuous injection from reals to circle is open map 
Let $f:\Bbb R\to S^1$ be continuous injective function (where $S^1$ is unit circle in $\Bbb R^2$). How to show that $f(a,b)$ is open for any $(a,b)\in \Bbb R$ where $a=-\infty$ and $b=\infty$ are possibilities?

I went like this: Maybe, we can consider $f$ as function from $\Bbb R\to [c,d)$, then it is strict monotone (since continuous and injective), then I can prove that $f(a,b)$ open. 
But, I can't justify why considering $f$ as function from $\Bbb R\to [c,d)$ is 
(Is there any direct approach to show that for $f:\Bbb R\to S^1$, f(a,b) is open?) 
 A: I think we can do this from scratch, without using covering maps or invoking invariance of domain: Let $(a,b)$ be a basis element for $\tau_{\mathbb R}$. Then, because $f$ is injective, $f((a,b))$ can not be onto $S^1.$ And then, since $f$ is continuous, and $(a,b)$ is connected, $f((a,b))$ is either an open, a half-open, or a closed arc. If the arc is open, we are done. If not, without loss of generality it is of the form $(zw]; z,w\in S^1.$ Now, let $(w-\epsilon,w+\epsilon)$ be an open arc containing $w$. There is a $a<c<b$ and a $\delta>0$ such that $f(c)=w,\ (c-\delta,c+\delta)\subseteq (a,b)$ and $f((c-\delta,c+\delta))\subseteq (w-\epsilon,w].$ But, $(c-\delta,c]$ and $(c,c+\delta)$ must be mapped by $f$ into disjoint arcs contained in $(w-\epsilon,w].$ But then, we get a contradiction because $f(c)\neq w:$ if so then there is an $\eta>0$ such that $f((c-\delta,c))\subseteq (\eta,w]$ such that $f(c)=w$ and $f((c,c+\delta)\subseteq (\epsilon_1,\eta)$ so $f$ is not continuous at $c$. 
A: The sort of argument the OP had in mind works with no problem if we know this (where $\Bbb T=\{z\in\Bbb C:|z|=1\}$):


Main Lemma. If $f:\Bbb R\to\Bbb T$ is continuous there exists a continuous $\theta:\Bbb R\to\Bbb R$ with $f(t)=e^{i\theta(t)}$.


(Note: When I wrote all this I was forgetting that $f$ is injective. That implies $f$ is not surjective, which makes the lemma completely trivial;  we an just let $\theta(t)=-i\log(f(t))$ for an appropriate branch of the logarithm in the plane. But the lemma is true without assuming $f$ is injective, and since this is an important fact we'll prove it for general continuous $f$.)
When I made a comment to this effect I was assuming that the lemma was something everybody knew. Evidently not. So:
Proof 1. This is clear from a little topology. The map $t\mapsto e^{it}$ is a covering map from $\Bbb R$ onto $\Bbb T$; now since $\Bbb R$ is simply connected any continuous map from $\Bbb R$ to $\Bbb T$ "lifts" through this covering map.
A slightly dodgier version of that argument in terms that may be more familiar to some readers:
Proof 2. Since $\Bbb R$ is simply connected and $f:\Bbb R\to\Bbb C$ is continuous and has no zero there exists a continuous branch of $\log(f)$ on $\Bbb R$.
Comment. Some readers are saying to themselves "we don't need no steenking covering maps". Yes you do need covering maps! For example, there are various results in elementary complex analysis that show simply connected domains have special properties. You do believe those results are things you need to know, and most of them are just covering-map arguments in disguise (see for example Complex Made Simple); they  make more sense if you look at them that way.
Or one can essentially include the covering-map argument without mentioning covering maps:
Proof 3. If $I\subset \Bbb R$ is an interval say $\theta:I\to\Bbb R$ is acceptable if $\theta $ is continuous and $f=e^{i\theta}$ on $I$.
First, it's clear that there exist local acceptable maps:


Lemma 2. If $t\in\Bbb R$ there exists $\delta>0$ such that there exists an acceptable  $\theta$ defined on  $I=(t-\delta,t+\delta)$.


Proof. Choose $\delta>0$ such that $f(I)\subset D(f(t),1)\subset \Bbb  C$. Let $L$ be a holomorphic branch of $\log z$ on $D(f(t),1)$, and let $\theta(s)=-iL(f(s))$.
And two local acceptable maps can be patched together:


Lemma 3. If $I_1$, $I_2$ are intervals and $\theta_j:I_j\to\Bbb R$ is acceptable then $\theta_2=\theta_1+2\pi k$ on $I_1\cap  I_2$; hence there exists an acceptable $\theta:I_1\cup I_2\to\Bbb R$.


Proof. $k=(\theta_2-\theta_1)/2\pi$ is a continuous integer-valued function on $I_1\cap I_2$, hence it is constant. Define $\theta:I_1\cup I_2\to\Bbb R$ by $$\theta(t)=\begin{cases}\theta_1(t),&(t\in I_1),
\\\theta_2(t)-2\pi k,&(t\in I_2).\end{cases}$$
Third Proof of the  Main Lemma. By compactness, Lemma 2 and Lemma  3 show that there exists an acceptable $\theta_k:[-k.k]\to\Bbb R$. Wlog $f(0)=1$ and $\theta_k(0)=0$; now it follows that $\theta_{k'}=\theta_k$ on $[-k'.k']\cap[-k,k]$...
Or if we assume $f$ is continuously differentiable we can give a proof using just calculus, no not-entirely-trivial topology:
Proof 4 (Main Lemma). Suppose $f$ is continuously differentiable. Assume wlog that $f(0)=1$. Define $$L(t)=\int_0^t\frac{f'(s)}{f(s)}\,ds.$$Then $f(0)=e^{L(0)}$ and $$\frac d{dt}\left( e^{-L(t)}f(t)\right)=0,$$so $e^{-L(t)}f(t)=1$.
(And $|e^L|=1$ shows that $L=i\theta$, $\theta\in\Bbb R$.)
Note. If you try to use the usual tricks (approximate identities/mollifiers) to derive the continuous case from the continuously differentiable case you may find it more convenient to prove an apparently stronger version of the lemma:


Main Lemma.02. If $f:\Bbb R\to\Bbb C\setminus\{0\}$ is continuous then there exists a continuous $L:\Bbb R\to\Bbb C$ with $f=e^L$.


Exercise. Explain why that version is only apparently stronger.
